What is the integral of (21)/(25+4r^2) ?

The integral of (21)/(25+4r^2) is 21/10 arctan(2/5 r)+C

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integral of (21)/(25+4r^2)

\int \frac{21}{25 + 4 r ^{2}} d r

\frac{21}{10} arctan (\frac{2}{5} r) + C

Solution steps

\int \frac{21}{25 + 4 r ^{2}} d r

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 21 \cdot \int \frac{1}{25 + 4 r ^{2}} d r

Apply Integral Substitution

= 21 \cdot \int \frac{1}{10 ( u ^{2} + 1 )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 21 \cdot \frac{1}{10} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= 21 \cdot \frac{1}{10} arctan (u)

Substitute back

u = \frac{2}{5} r

= 21 \cdot \frac{1}{10} arctan (\frac{2}{5} r)

Simplify

21 \cdot \frac{1}{10} arctan (\frac{2}{5} r) : \frac{21}{10} arctan (\frac{2}{5} r)

= \frac{21}{10} arctan (\frac{2}{5} r)

Add a constant to the solution

= \frac{21}{10} arctan (\frac{2}{5} r) + C

\int x^{2} \cdot cos (n x) d x

\frac{d}{d x} (\frac{2 x}{x - 4})

\frac{d}{d x} (\frac{2 x}{x ^{2} + 9})

in v erse l a pl a ce \frac{3 s - 15}{2 s ^{2} - 4 s + 10}

t an g e n t f (x) = \frac{2}{x + 7}, (- 8, - 2)

What is the integral of (21)/(25+4r^2) ?
The integral of (21)/(25+4r^2) is 21/10 arctan(2/5 r)+C